Answer:
I cheated and asked Richard Lyons this question (or at least, the reformulation of the problem, conjecturing that (G,c) is nonsplitting for an involution c with <c> generating G if and only if there exists an odd A such that G/A = Z/2). His response:
Good question! This is a famous (in my circles) theorem - the Glauberman Z^*-Theorem. (Z^*(G) is the preimage of the exponent 2 subgroup of the center of G/O(G), and O(G)=largest normal subgroup of G of odd order.)
Z^*-Theorem: If c is an involution of G then c\in Z^*(G) iff [c,g] has odd order for all g\in G iff for any Sylow 2-subgroup S of G containing c, c is the unique G-conjugate of itself in S.
The last property is absolutely fundamental for CFSG. The proof uses modular character theory for p=2. Attempts to do it with simpler tools have failed.
George Glauberman, Central Elements in Core-free Groups, Journal of Algebra 4, 1966, 403-420.
Older Remarks:
Comment 1: Suppose that P = Z/2+Z/2 is a 2-Sylow. If x lies in P, then P clearly centralizes x, and thus the order of <x> divides #G/P, and is thus odd. By a theorem of Frobenius, G has an odd number of elements of order 2, and thus we see it has an odd number of conjugacy classes of elements of order 2. Yet, by the Sylow theorems, every element of order 2 is conjugate to an element of P. If c lies in P, then by nonsplitting, it is unique in its G-conjugacy class in P. Thus there must be exactly three conjugacy classes of elements of order 2, and thus no element of P is G-conjugate. By a correct application of Frobenius' normal complement theorem, we deduce that G admits a normal subgroup A such that G/A \sim P. Yet <c> generates G, and thus the image of <c> generates G/A. Yet G/A is abelian and non-cyclic, a contradiction.
Comment 2: Suppose that A is a group of order coprime to p such that p | #Aut(A).
Let G be the semidirect product which sits inside the sequence:
1 ---> A ---> G --(phi)--> Z/pZ --> 0;
Let c be (any) element of order p which maps to 1 in Z/pZ. If c is conjugate to
c^j, then phi(c) = phi(c^j). Hence c is not conjugate to any power of itself.
Let H be a subgroup of G containing c (or a conjugate of c, the same argument applies). The element c generates a p-sylow P
of H (and of G). It suffices to show that if gcg^-1 lies in H, then it is conjugate to c inside H. Note that gPg^-1 is a p-Sylow of H.
Since all p-Sylows of H are conjugate, there exists an h such that gPg^-1 = hPh^-1, and thus h c^j h^-1 = gcg^-1. Yet we have seen that c^j is not conjugate to c inside G unless j = 1. Thus gcg^-1 = hch^-1 is conjugate to c inside H.
I just noticed that you wanted <c> to generate G. It's not immediately clear (to me) what condition on A one needs to impose to ensure this. Something like the automorphism has to be "sufficiently mixing". At the very worst, I guess, the group G' generated by <c> still has the property, by the same argument.
This works more generally if p || G and no element of order p is conjugate to a power of itself. (I think you know this already if p = 2.)
The case where the p-Sylow is not cyclic is probably trickier.
Examples: A = (Z/2Z)+(Z/2Z), p = 3. (This is A_4).
A = Quaternion Group, p = 3. (This is GL_2(F_3) = ~A_4, ~ = central extension).
A = M^37, M = monster group, p = 37.
Best Answer
This might help: If $u$ and $v$ are both close enough to scalars in the operator norm, and $\varepsilon$ is small enough, then no matter how large $k$ is, if the group generated by $u^{\prime}$ and $v^{\prime}$ is finite, then it still has to be Abelian. The reasoning is something like this: I use the operator norm : given a unitary matrix $M$, there is a unique scalar $\lambda$ ( in the interior of the unit disk unless $M$ itself is a scalar matrix) with $\|M- \lambda I\|$ minimal, and this only depends on the spectrum of $M$. In particular, we can only have $\|M -\lambda I\| < \frac{1}{2}$ for a scalar $\lambda$ if the eigenvalues of $M$ all lie on an arc of length less than $\frac{\pi}{3}$ on the unit circle.It is essentially a Theorem of Frobenius that if $G$ is a finite group of unitary matrices, then the subgroup $H$ generated by the matrices at distance less than $\frac{1}{2}$ from a scalar is an Abelian normal subgroup. If $u$ and $v$ are close enough to scalars, and $\varepsilon$ is small enough then we can make both $\|u^{\prime} - \lambda I\|$ and $\| v^{\prime} - \mu I\|$ less than $\frac{1}{2}$ for suitable scalars $\lambda,\mu$. So if they generate a finite group, it must be Abelian. Note added: I will add a little more detail. One proof of Jordan's theorem uses a compactness argument and a kind of contraction Lemma. Let $[a,b]$ denote $a^{-1}b^{-1}ab$. If $a$ and $b$ are unitary, then $$\| I-[a,b] \| = \|I - a^{-1}b^{-1}ab \| = \|ab-ba \|$$ $$ = \| (a-I)(b-I) - (b-I)(a-I)\| \leq 2 \|a-I\| \|b-I\|$$. Hence if $\| I-a\| < \frac{1}{2}$, then $\|I- [a,b] \| < \|I-b\|$ for all $b.$ Hence in a finite group $G$ of unitary matrices, let $H$ be the subgroup generated by those elements of $G$ with $\|I-h\|< \frac{1}{2}$. For any $h \in H$ with $\|I - h \| < \frac{1}{2}$, we have $[g,h,h, \ldots ,h] = 1$ for all $g \in G$. By a theorem of Baer, $H$ is nilpotent. With a little more work, you actually get $H$ Abelian. (This is a kind of hybrid of arguments of Frobenius and a little more finite group theory. More or less the same line of reasoning led to a similar theorem by Zassenhaus on discrete groups, and this led to the idea of a Zassenhaus neighbourhood in a Lie group). Note that if $x,y$ are in different cosets of $H$ in $G$, we must have $\|x-y\| \geq \frac{1}{2}$, so the compactness of the unit ball of $M_{n}(\mathbb{C})$ gives a fixed bound on $[G:H]$. But the same argument works when $\| a - \lambda I \| < \frac{1}{2}$ for some scalar $\lambda$, using $$\|ab - ba \| = \| (a-\lambda I)(b-\mu I) - (b-\mu I)(a-\lambda I)\|$$ for any scalars $\lambda,\mu$, and replacing $H$ by the group generated by elements at distance less than $\frac{1}{2}$ from any scalar. (Added in response to comment below): You do need a little argument to get from $H$ nilpotent to $H$ Abelian, but the statements above are all accurate. It is important to remember that $H$ is generated by elements within distance $\frac{1}{2}$ of a scalar. Any such element has all its eigenvalues on an arc of length less than $\frac{\pi}{3}$ on $S^{1}$. No subset of the "multiset" of eigenvalues of such an element has zero sum (it's clear eg, if you rotate so that all eigenvalues have positive real part). Let $\chi$ be the character of the given representation of $H$. Let $\mu$ be an irreducible constituent of $\chi$. By the remark above, if $\|h -\alpha I \| < \frac{1}{2}$ for some scalar $\alpha$, then $\mu(h) \neq 0.$ But if $\mu$ is not linear, it is induced from a character of a maximal subgroup $M$, which is normal as $H$ is nilpotent. But then $\mu$ vanishes on all elements of $H$ which are outside $M$. Hence the elements at distance less than $\frac{1}{2}$ from a scalar must all lie within $M$. But since $M$ is proper, and such elements generate $H$, this is a contradiction. Hence $\mu$ must be linear after all. Since $\mu$ was arbitrary, $H$ is Abelian.